Using an integral transformation, the Riccati transformation, and iteration, we obtain sufficient conditions for all solutions to be oscillatory

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

OSCILLATION CRITERIA FOR EVEN-ORDER NONLINEAR NEUTRAL DIFFERENCE EQUATIONS WITH CONTINUOUS

VARIABLES

SHUHUI WU, PARGAT SINGH CALAY, ZHANYUAN HOU

Abstract. In this article, we study the oscillatory behavior of solutions to even-order nonlinear neutral difference equations of the form

∆^m_τ(x(t)−px(t−r)) +f(t, x(g(t))) = 0.

Using an integral transformation, the Riccati transformation, and iteration, we obtain sufficient conditions for all solutions to be oscillatory. Examples are also given to illustrate the obtained criteria.

1. Introduction

Difference equations have attracted a great deal of attention of researchers in mathematical, biological, physical sciences, and economy. This is specially due to the applications in various problems of biology, physics, economy, and so on. The topics studied for oscillation of the solutions have been investigated intensively and the references [1]–[17] are just a few examples.

In this article, we study even-order nonlinear neutral difference equations with continuous variable of the form

∆^m_τ(x(t)−px(t−r)) +f(t, x(g(t))) = 0, (1.1) wheremis an even integerm≥4,p≥0,τ andrare positive constants, ∆τx(t) = x(t+τ)−x(t), 0< g(t)< t,g ∈C¹([t0,∞),R⁺), g⁰(t)>0, and f ∈C([t0,∞)× R,R). Throughout this article we assume that

g(t+τ)≥g(t) +τ fort≥t0, (1.2)

f(t, u)/u≥q(t)>0 foru6= 0 and q∈C(R,R⁺). (1.3) Lett^∗₀= min{g(t₀), t₀−r}andI₀= [t^∗₀, t₀]. A functionxis called thesolution of (1.3) withx(t) =ϕ(t) fort∈I₀ andϕ∈C(I₀,R) if it satisfies (1.3) fort≥t₀.

A solution xis called oscillatory if it has arbitrarily large zeros; otherwise it is non-oscillatory. xis eventually positive if there exists t1 ≥t0, such thatx(t)>0 for allt≥t1. Eventually negative definite is defined similarly.

This article is organized as follows. The main results is stated in section 2 and leave the proofs for section 5. Examples will be presented in section 3 to

2010Mathematics Subject Classification. 39A10, 39A21.

Key words and phrases. Neutral difference equations; oscillation; continuous variable.

c

2015 Texas State University - San Marcos.

Submitted November 2, 2014. Published May 7, 2015.

1

demonstrate the application of the obtained results. In section 4, some lemmas will be given to be used in the proofs of the main results.

2. Statement of main results

The assumptions ongguarantee the existence and differentiability of its inverse g⁻¹. Let

¯

q(t) =α min

t≤s≤t+mτ{q(s)}

min

g(t)≤s≤g(t)+mτ{(g⁻¹(s))⁰}m

, (2.1)

where 0 < α < 1. The function ¯q will play an important role in the oscillatory criteria for (1.3). Throughout this article, we use the symbol dae to denote the smallest integer not less thana.

Theorem 2.1. Assume that for some t1≥t0,

∞

X

i=0

¯

q(t1+iτ) =∞. (2.2)

Then for every solution x(t)of (1.1), is either oscillatory, or for anyT ≥t₀ there exists at₂> T such that |x(t₂)| ≤p|x(t₂−r)|.

Theorem 2.2. In addition to (2.2), we assume that 0 < p < 1 and there is a positive integerk₀ and at₁≥t₀ satisfyingm₁(n) =d(g(t₁+nτ)−t₁+k₀r)/τe ≤n for all large enough n. Moreover, assume that there is a sequence{n_k} → ∞such

that n_k

X

i=m1(nk)

¯

q(t1+iτ)≥ p^k0(1−p)

1−p^k0 (2.3)

for allk large enough. Then for every solution x(t)of (1.1), either x(t)orx(t)− px(t−r)or both are oscillatory.

Corollary 2.3. In addition to (2.2), we assume that 0 < p < 1 and there is a positive integerk0 and at1≥t0 satisfyingm1(n) =d(g(t1+nτ)−t1+k0r)/τe ≤n for all sufficiently largen. Moreover, we assume that there is a sequence{nk} → ∞ such that

n_k

X

i=m₁(n_k)

(i−m1+ 1)¯q(t1+iτ)≥ p^k0(1−p)

1−p^k0 (2.4)

for allk large enough. Then for every solution x(t)of (1.1), either x(t)orx(t)− px(t−r)or both are oscillatory.

Remark 2.4. Note that the requirement (2.4) for ¯q(t) is weaker than (2.3) since (i−m1+ 1)≥1 holds in (2.4).

Corollary 2.5. In addition to (2.2), we assume that0< p <1 and that there is a positive integerk0 and at1≥t0 satisfyingm1(n) =d(g(t1+nτ)−t1+k0r)/τe ≤n for all sufficiently large n. Moreover, assume that there is a sequence{nk} → ∞, and an integerl,1≤l < msuch that

1 l!

n_k

X

i=m₁(n_k)

(i−m1+ 1)(i−m1+ 2). . .(i−m1+l)¯q(t1+iτ)≥ p^k0(1−p) 1−p^k0 (2.5) for allk large enough. Then for every solution x(t)of (1.1), either x(t)orx(t)− px(t−r)or both are oscillatory.

Remark 2.6. Note that (2.5) coincides with (2.4) forl= 1. Forl >1, (i−m₁+ 1)(i−m₁+ 2). . .(i−m₁+l)

l! ≥i−m1+ 1≥1.

Thus, (2.5) is weaker than (2.4) and (2.3).

Theorem 2.7. In addition to (2.2), we assume that p = 1 and that there is a positive integerk0 and at1≥t0 satisfyingm1(n) =d(g(t1+nτ)−t1+k0r)/τe ≤n for all sufficiently large n. Moreover assume that there is a sequence {nk} → ∞ such that

n(k)

X

i=m₁(n_k)

¯

q(t₁+iτ)≥ 1 k0

(2.6) forklarge enough. Then for every solutionx(t)of (1.1), eitherx(t)orx(t)−x(t−r) or both are oscillatory.

Corollary 2.8. In addition to (2.2), we assume that p = 1 and that there is a positive integerk0 and at1≥t0satisfyingm1(nk) =d(g(t1+nτ)−t1+k0r)/τe ≤n for all sufficiently large n. Moreover assume that there is a sequence {nk} → ∞ such that

n_k

X

i=m1(n_k)

(i−m₁+ 1)¯q(t₁+iτ)≥ 1

k₀ (2.7)

forklarge enough. Then for every solutionx(t)of (1.1), eitherx(t)orx(t)−x(t−r) or both are oscillatory.

Corollary 2.9. In addition to (2.2), we assume that p = 1 and that there is a positive integerk0 and at1≥t0satisfyingm1(nk) =d(g(t1+nτ)−t1+k0r)/τe ≤n for all sufficiently largen. Moreover we assume that there is a sequence{nk} → ∞ and an integerl,1≤l≤m−1 such that

1 l!

n_k

X

i=m1(nk)

(i−m1+ 1)(i−m1+ 2). . .(i−m1+l)¯q(t1+iτ)≥ 1

k₀ (2.8) for k large enough. Then, for every solution x(t) of (1.1), either x(t) or x(t)− x(t−r)or both are oscillatory.

Theorem 2.10. In addition to (2.2), we assume that p > 1 and that there is a positive integerk0 and at1≥t0satisfyingm1(nk) =d(g(t1+nτ)−t1+k0r)/τe ≤n for all large enough n. Moreover assume that there is a sequence {nk} → ∞such

that n_k

X

i=m1(nk)

¯

q(t1+iτ)≥ p^k0(1−p)

1−p^k0 (2.9)

holds for allklarge enough. Then, for every bounded solution x(t)of (1.1), either x(t)orx(t)−px(t−r)or both are oscillatory.

Corollary 2.11. In addition to (2.2), we assume that p >1 and that there is a positive integerk₀ and at₁≥t₀ satisfyingm₁(n) =d(g(t1+nτ)−t₁+k₀r)/τe ≤n for all large enough n. Moreover assume that there is a sequence {n_k} → ∞such

that n_k

X

i=m₁(n_k)

(i−m1+ 1)¯q(t1+iτ)≥ p^k0(1−p)

1−p^k0 (2.10)

holds for all k large enough. Then for every bounded solutionx(t)of (1.1), either x(t)orx(t)−px(t−r)or both are oscillatory.

Corollary 2.12. In addition to (2.2), we assume that p >1 and that there is a positive integerk0 and at1≥t0 satisfyingm1(n) =d(g(t1+nτ)−t1+k0r)/τe ≤n for all sufficiently large n. Moreover assume that there is a sequence {nk} → ∞ and an integerl,1≤l < m, such that

1 l!

n_k

X

i=m₁(n_k)

(i−m1+ 1)(i−m1+ 2). . .(i−m1+l) ¯q(t1+iτ)≥ p^k0(1−p) 1−p^k0 (2.11) holds for all k large enough. Then, for every bounded solution x(t) of equation (1.1), eitherx(t)orx(t)−px(t−r)or both are oscillatory.

Remark 2.13. Note that

(i−m₁+ 1)(i−m₁+ 2). . .(i−m₁+l−1)

(l−1)! ≥i−m1+ 1≥1

holds. Thus, (2.11) is weaker that (2.10) and (2.9).

Corollary 2.14. In addition to (2.2), we assume that p >1 and that there is a positive integerk₀ and at₁≥t₀ satisfyingm₁(n) =d(g(t₁+nτ)−t₁+k₀r)/τe ≤n for all sufficiently large n. Moreover assume that there is a sequence {n_k} → ∞ such that

1 (n−1)!

n_k

X

i=m1(nk)

(i−m1+n−1)!

(i−m₁)! q(t¯ 1+iτ)≥ p^k0(1−p)

1−p^k0 (2.12) holds for allklarge enough. Then, for every bounded solution x(t)of (1.1), either x(t)orx(t)−px(t−r)or both are oscillatory.

3. Examples

Three illustrating examples are given here to demonstrate the applications of the obtained oscillatory criteria.

Example 3.1. Consider the linear difference equation

∆²ⁿ_τ (x(t)−px(t−r)) +1

tx(t− σ

1 +βt) = 0 (3.1)

for t > 0, where n is a positive integer, p ≥0, β ≥0, the constants r, τ and σ are positive. Viewing (3.1) as (1.1), we haveq(t) = 1/tandg(t) =t−σ/(1 +βt).

Then, according to (2.1), ¯q(t) =α/(t+ 2nτ) forβ= 0 and

¯

q(t) = α t+ 2nτ

1− σβ

(1 +βt)²+σβ ²ⁿ

forβ >0. Since ¯q2n(t)≥α⁰/(t+ 2nτ) for someα⁰ >0 and allt≥0, ¯q2n satisfies (2.2) with t1 = 0. By Theorem 2.1, for every solution x(t) of (3.1), either x(t) is oscillatory or for anyT ≥t0 there exists a t2> T such that |x(t2)|< p|x(t2−r)|.

In particular, whenp= 0, every solution of (3.1) is oscillatory.

Example 3.2. Consider the difference equation

∆²ⁿ_π (x(t)−px(t−π)) + 8x(t−π) + 8σ

1 +t²x³(t−π) = 0, (3.2)

where σ ≥ 0 is a constant. Regarding (3.2) as (1.1), we have τ = π, r = π, g(t) =t−π andq(t) = 8. Then, for someα∈(0,1),q¯2n= 8αby (2.1)so (2.2)is satisfied. Forp= 1,k0= 1 andt1=t, we have ml=l and

l

X

s=ml

(s+ 1−ml)¯q2n(t1+sτ) = 8α >1 = 1 k0

if α >1/8. Also we have

l

X

s=ml

(s+ 1−m_l)¯q_2n(t₁+sτ) = 8α > p= (1−p)p^k0 1−p^k0

if p ∈ (0,1)∪(1,8) and α > p/8. According to Theorems 2.2 and 2.7, for every solution x(t) of (3.2), either x(t) or x(t)−px(t−r) is oscillatory if 0 < p ≤1.

Furthermore, by Theorem 2.10, for every bounded solutionx(t)of (3.2), eitherx(t) orx(t)−px(t−r)is oscillatory if1< p <8.

Example 3.3. Consider the difference equation

∆²ⁿ_τ (x(t)−x(t−r)) + 2²ⁿ⁺¹x(t−3) = 0, (3.3) whereτ andrare positive odd integers. Viewing (3.3) as (1.1), we haveg(t) =t−3 and q(t) = 2²ⁿ⁺¹. Then, for some α ∈ (0,1), ¯q2n = α2²ⁿ⁺¹ by (2.1) so (2.2) is satisfied. Forp= 1,k0= 1 andt1=t, we haveml=land

l

X

s=m_l

(s+ 1−ml)¯q2n(t1+sτ) =α2²ⁿ⁺¹>1 = 1 k₀

ifα >2^−(2n+1). According to Theorem 2.7, for every solutionx(t) of (3.3), either x(t) orx(t)−x(t−r) is oscillatory.

4. Related lemmas

In this section, we present the lemmas which will be needed in the proofs of the main results. The following lemma can be found in [1, page 31].

Lemma 4.1. Letu(k)be defined onN(a), wherea∈N, andu(k)>0with∆^mu(k) of constant sign onN(a)for any positive integermand not identically zero. Then, there exists an integer h, 0 ≤h≤m, with m+h odd for ∆^mu(k) ≤0 or m+h even for∆^mu(k)≥0 such that

(i) h≤m−1 implies(−1)^h+i∆ⁱu(k)>0for all k∈N(a),h≤i≤m−1, (ii) h≥1implies ∆ⁱu(k)>0 for allk∈N(a),1≤i≤h−1.

By applying the above result to the difference with continuous variables, we have the following lemma.

Lemma 4.2. Let y(t) be defined on [t₀,+∞) where t₀ ∈ R, and y(t) > 0 with

∆^m_τ y(t)of constant sign on[t₀,+∞)for any positive integerm and not identically zero. Then, there exists an integer h, 0 ≤h≤m, withm+h odd for∆^m_τy(t)≤ 0 or m+h even for ∆^m_τy(t)≥0 such that

(i) h≤m−1 implies(−1)^h+i∆ⁱ_τy(t)>0 for all t∈[t0,∞),h≤i≤m−1, (ii) h≥1implies ∆ⁱ_τy(t)>0 for allt∈[t0,+∞),1≤i≤h−1.

Proof. Let t1 be any constant real number in [t0,+∞). For this fixed t1, by the assumption, we havey(t1+kτ) defined for anyk∈ {0,1, . . .}, and y(t1+kτ)>0 with ∆^m_τy(t1 +kτ) of constant sign for any k ∈ {0,1, . . .} and for any positive integermand not identically zero. Thus, by Lemma 4.1, the conclusion holds with the replacement of t byt₁+kτ for allk ∈ N. Since t₁ ∈ [t₀,∞) is arbitrary, we can see that the conclusion holds fort∈[t₀,+∞).

Lemma 4.3([1, page 289]). Let y(t)be an mtimes differentiable function onR+

of constant sign satisfying y^(m)(t) 6≡0 and y^(m)(t)y(t)≤ 0 on [t1,∞). Then the following statements hold.

(i) There exists a t2 ≥t1 such that the functionsy^(j)(t),j = 1,2, . . . , m−1, are of constant sign on[t₂,∞).

(ii) There exists an integerk < m which is odd (even) whenm is even (odd), such that

y(t)y^(j)(t)>0 forj= 0,1, . . . , k, t≥t₂, (−1)^m+j+1y(t)y^(j)(t)>0 forj=k+ 1, . . . , m, t≥t₂.

Lemma 4.4 ([5, page 289]). Assume that y(t), y⁰(t), . . . , y^(m−1)(t) are absolutely continuous and of constant sign on the interval(t0,∞), and assumey^(m)(t)y(t)≥0.

Then either y^(k)(t)y(t) ≥ 0, k = 0,1, . . . , m or there exists an integer l,0 ≤ l ≤ m−2, which is even (odd) whenm is even (odd), such that

y^(k)(t)y(t)≥0, fork= 0,1, . . . , l, (−1)^m+ky^(k)(t)y(t)≥0, fork=l+ 1, . . . , m.

Lemma 4.5. Assume thatx(t)is an eventually positive (negative) solution of (1.1) such thaty(t) =x(t)−px(t−r)>0 (<0) eventually. Then∆_τy(t)>0 (<0)and

∆^m−1_τ y(t)>0 (<0) hold eventually.

Proof. Supposex(t)>0 and y(t)>0 hold eventually. Due tog(t)< t, g⁰(t)>0 and (1.2), there exists a t₁> t₀such that x(g(t))>0 for all t≥t₁. Further, (1.1) becomes

∆^m_τ y(t) +f(t, x(g(t))) = 0.

According to (1.3),f(t, x(g(t)))≥q(t)x(g(t))>0 fort≥t1hold. Therefore,

∆^m_τy(t)≤ −q(t)x(g(t))<0 (4.1) for all large enough t, namely, ∆^m_τy(t) < 0 eventually. By Lemma 4.2, h could be odd with 1 ≤ h ≤ m−1. For all cases, we could obtain ∆_τy(t) > 0 and

∆^m−1_τ y(t) > 0 eventually. If x(t) < 0 andy(t) < 0 hold eventually, then (4.1) becomes ∆^m_τ y(t) ≥ −q(t)x(g(t)) >0. Applying Lemma 4.2 to −y(t), we obtain

∆τy(t)<0 and ∆^m−1_τ y(t)<0.

Lemma 4.6. Let the hypothesis of Lemma 4.5 be satisfied. Moreover, let q(t)¯ be defined by (2.1). Set

u(t) = Z t+τ

t

dt1

Z t1+τ

t1

dt2. . .

Z tm−2+τ

tm−2

dt_m−1

Z tm−1+τ

tm−1

y(θ)dθ.

Then u satisfies u^(m)(t) = ∆^m_τy(t) < 0 (> 0), u(t) > 0 (<0), u⁰(t) > 0 (< 0), u^(m−1)(t)>0 (<0),∆^m−1_τ u(t)>0 (<0), and

∆^m_τu(t) + ¯q(t)u(g(t)−kr)

k

X

i=0

pⁱ≤0 (≥0) for each fixed numberk and for all large enought.

Proof. Supposex(t)>0 andy(t)>0 hold eventually. According to the definition ofu(t) and (4.1), we can see thatu(t)>0,u^(m)(t) = ∆^m_τ y(t)<0 and

∆^m_τy(t) +q(t)x(g(t))≤0 (4.2)

for sufficiently larget. Taking into account the definition ofy(t), we have

∆^m_τ y(t) +q(t)(y(g(t)) +px(g(t)−r))≤0.

By repeating the above processktimes, we deduce

∆^m_τy(t) +q(t)

k

X

i=0

pⁱy(g(t)−ir) +q(t)p^k+1x(g(t)−(k+ 1)r)≤0.

Therefore, sinceq(t)p^k+1x(g(t)−(k+ 1)r)≥0, it follows that

∆^m_τy(t) +q(t)

k

X

i=0

pⁱy(g(t)−ir)≤0.

Furthermore,

u^(m)(t) +q(t)

k

X

i=0

pⁱy(g(t)−ir)≤0. (4.3)

Then, for large enought, the assumptions ong andqgive Z t+τ

t

ds1

Z s₁+τ

s₁

dsm−2. . .

Z sm−2+τ

sm−2

dsm−1

Z sm−1+τ

sm−1

y(g(θ)−ir)q(θ)dθ

≥ min

t≤l≤t+mτ{q(l)}

Z t+τ

t

ds1

Z s₁+τ

s1

ds_m−2. . .

×

Z sm−2+τ

sm−2

ds_m−1

Z sm−1+τ

sm−1

y(g(θ)−ir)dθ

≥ min

t≤l≤t+mτ{q(l)}

Z g(t+τ)

g(t)

(g⁻¹(s1))⁰ds1

Z g(g⁻¹(s₁)+τ)

s₁

(g⁻¹(s2))⁰ds2. . .

×

Z g(g⁻¹(s_m−2)+τ)

sm−2

(g⁻¹(s_m−1))⁰ds_m−1

Z g(g⁻¹(s_m−1)+τ)

sm−1

y(θ−ir)(g⁻¹(θ))⁰dθ

≥ min

t≤l≤t+mτ{q(l)}

min

g(t)≤s≤g(t)+mτ(g⁻¹(s))^0mZ g(t)+τ g(t)

ds1

Z s₁+τ

s1

ds2. . .

×

Z sm−2+τ

sm−2

ds_m−1

Z sm−1+τ

sm−1

y(θ−ir)dθ

≥ min

t≤l≤t+mτ{q(l)}

min

g(t)≤s≤g(t)+mτ(g⁻¹(s))⁰min)^mu(g(t)−ir)

≥q(t)u(g(t)¯ −ir).

Thus, integration on both sides of (4.3) gives

∆^m_τ u(t) + ¯q(t)

k

X

i=0

pⁱu(g(t)−ir)≤0. (4.4)

According to the definition ofu(t), the equality u⁰(t) =

Z t+τ

t

dt₂ Z t2+τ

t₂

dt₃. . .

Z tm−2+τ

tm−2

dt_m−1

Z tm−1+τ

tm−1

∆_τy(θ)dθ holds. Then it follows from Lemma 4.5 thatu⁰(t)>0. Similarly, we have

u^(m−1)(t) = Z t+τ

t

∆^m−1_τ y(θ)dθ sou^(m−1)(t)>0 from Lemma 4.5. Hence,

∆^m−1_τ u(t) = Z t+τ

t

dt₁ Z t1+τ

t1

dt₂. . .

Z tm−2+τ

tm−2

u^(m−1)(θ)dθ >0.

Further, (4.4) implies

∆^m_τu(t) + ¯q(t)u(g(t)−kr)

k

X

i=0

pⁱ≤0

for each fixed natural numberkand for all large enought. Ifx(t)<0 andy(t)<0 hold eventually, thenu(t)<0,u^(m)(t) = ∆^m_τ y(t)>0 and ∆^m_τ y(t) +q(t)x(g(t))≥0 for large enought. Moreover, (4.3) becomes

u^(m)(t) +q(t)

k

X

i=0

pⁱy(g(t)−ir)≥0 and (4.4) becomes

∆^m_τ u(t) + ¯q(t)

k

X

i=0

pⁱu(g(t)−ir)≥0.

Thatu⁰(t)<0 andu^(m−1)(t)<0 follow from ∆_τy(t)<0 and ∆^m−1_τ y(t)<0. Then

∆^m−1_τ u(t)<0 follows from the integration ofu^(m−1)(t). Since u(t) is decreasing, eachu(g(t)−ir) can be replaced by u(g(t)−kr) in the above inequality.

5. Proofs of the main results

Proof of Theorem 2.1. Letx(t) be a solution of (1.1) satisfyingx(t)>0 andx(t)−

px(t−r)>0 for all larget. Lety(t) be as in Lemma 4.5 and u(t) be as in Lemma 4.6. Furthermore, for any positive integerk, we have

∆^m_τu(t) + ¯q(t)u(g(t)−kr)

k

X

i=0

pⁱ≤0, whereu(g(t)−kr)>0. Define the Riccati transformation by

v(t) = ∆^m−1_τ u(t) u(g(t)−kr). Notice thatv(t)>0. Moreover we deduce

∆_τv(t) =v(t+τ)−v(t)

= ∆^m−1_τ u(t+τ)

u(g(t+τ)−kr)− ∆^m−1_τ u(t) u(g(t)−kr)

= u(g(t)−kr)∆^m−1_τ u(t+τ)−u(g(t+τ)−kr)∆^m−1_τ u(t) u(g(t+τ)−kr)u(g(t)−kr)

= u(g(t)−kr)∆^m−1_τ u(t+τ) +u(g(t+τ)−kr)(∆^m_τu(t)−∆^m−1_τ u(t+τ)) u(g(t+τ)−kr)u(g(t)−kr)

≤ ∆^m_τu(t)

u(g(t)−kr)−∆^m−1_τ u(t+τ)∆τu(g(t)−kr) u(g(t)−kr)u(g(t+τ)−kr)

≤ −q(t)¯

k

X

i=0

pⁱ−v(t+τ)∆τu(g(t)−kr) u(g(t)−kr)

≤ −q(t)¯

k

X

i=0

pⁱ.

Therefore, there exists at₁> t₀ such that

∆_τv(t₁+jτ) + ¯q(t₁+jτ)

k

X

i=0

pⁱ≤0. (5.1)

Summing both sides of (5.1) from 0 ton, we have v(t₁+ (n+ 1)τ)−v(t₁) +

k

X

i=0

pⁱ

n

X

j=0

¯

q(t₁+jτ)≤0.

Thus

k

X

i=0

pⁱ

n

X

j=0

¯

q(t1+jτ)< v(t1)<∞,

which leads to a contradiction to (2.2). If x(t) is a solution of (1.1) satisfying x(t)<0 and y(t)<0 eventually, from Lemmas 4.5 and 4.6, the above argument aboutv(t) is still valid and also leads to a contradiction. Therefore, the conclusion

of the theorem holds.

Proof of Theorem 2.2. According to Theorem 2.1, if (2.2) holds, we have that every solutionx(t) of (1.1) is either oscillatory or for anyT ≥t0, there exists onet2> T such that|x(t2)| ≤p|x(t2−r)|.

Assume that (1.1) has an eventually positive solution x(t) such that y(t) = x(t)−px(t−τ) is not oscillatory. Then from Theorem 2.1, we deduce thaty(t)<0 for all large enought. Letz(t) =−y(t). Therefore,z(t)>0 and

∆^m_τz(t)−f(t, x(g(t))) = 0.

Moreover,

∆^m_τz(t)≥q(t)x(g(t))>0 so

∆^m_τz(t)−q(t)x(g(t))≥0. (5.2)

Forz(t), according to Lemma 4.2,his even. So ∆ⁱ_τz(t)>0 for all even number iwith 2≤i≤m−2, and|∆^j_τz(t)|>0 for all odd numberj with 1≤j≤m−1.

We show that ∆τz(t)< 0. Indeed, if ∆τz(t)> 0, then, since ∆²_τz(t) >0, we may assume ∆τz(t1+kτ)> l >0 for a large enought1 and allk∈N. Then

d

X

i=0

∆_τz(t₁+iτ) =z(t₁+ (d+ 1)τ)−z(t₁)≥(d+ 1)l.

Letd→ ∞, thenz(t1+ (d+ 1)τ)→+∞. We have lim_t→∞x(t) = 0 by repeating x(t)< px(t−r) for 0< p <1. Thus, by the definition ofz(t), we have lim_t→∞z(t) = 0 which contradictsz(t1+dτ)→+∞asd→ ∞. Thus, ∆τz(t)<0.

So, according to Lemma 4.2 again,h= 0. Thus, ∆ⁱ_τz(t)>0 for all even number iwith 2≤i≤m−2, and ∆^j_τz(t)<0 for all odd numberj with 1≤j≤m−1.

Noticex(t) = (x(t+r) +z(t+r))/p. Hence, from (5.2), it follows that

∆^m_τz(t)−q(t)

p z(g(t) +r)−q(t)

p x(g(t) +r)≥0, and further

∆^m_τz(t)−q(t)

k

X

i=1

1

pⁱ z(g(t) +ir)−q(t)

p^k x(g(t) +kr)≥0.

So,

∆^m_τ z(t)−q(t)

k

X

i=1

1

pⁱz(g(t) +ir)>0 (5.3)

sincex(g(t) +kr)>0. Let u(t) =

Z τ

0

ds₁ Z s1+τ

s₁

ds₂. . .

Z sm−2+τ

sm−2

ds_m−1

Z t+sm−1+τ

t+sm−1

z(θ)dθ.

Then we haveu^(m)(t)>0 andu(t)>0. Since u⁰(t) =

Z τ

0

ds1

Z s₁+τ

s1

ds2. . .

Z sm−2+τ

sm−2

∆τz(t+s_m−1)ds_m−1,

then ∆τz(t)<0 impliesu⁰(t)<0. Moreover,u⁽ⁱ⁾(t)>0 for all even numberiwith 2≤i≤m−2, andu^(j)(t)<0 for all odd numberj with 1≤j≤m−1.

Integrating (5.3) and from the proof of Lemma 4.6 replacing y(t) by z(t), we have

∆^m_τu(t)−q(t)¯

k

X

i=1

1

pⁱu(g(t) +ir)>0, which leads to

∆^m_τ u(t)−q(t)u(g(t) +¯ kr)

k

X

i=1

1 pⁱ >0.

Due toPk

i=11/pⁱ = (1−p^k)/(p^k(1−p)), we deduce that

∆^m_τu(t)≥ 1−p^k

p^k(1−p)q(t)¯ u(g(t) +kr)>0.

Replacingkbyk₀ andtbyt₁+iτ in the above inequalities yield

∆^m_τu(t₁+iτ)≥ 1−p^k0

p^k0(1−p)q(t¯ ₁+iτ)u(g(t₁+iτ) +k₀r).

Summing up both sides of the above inequality forifromstonand sinceu⁰(t)<0, we have

∆^m−1_τ u(t1+(n+1)τ)−∆^m−1_τ u(t1+sτ)≥ 1−p^k0

p^k0(1−p)u(g(t1+nτ)+k0r)

n

X

i=s

¯

q(t1+iτ), which implies

−∆^m−1_τ u(t1+sτ)> 1−p^k0

p^k0(1−p)u(g(t1+nτ) +k0r)

n

X

i=s

¯

q(t1+iτ) (5.4) due to ∆^m−1_τ u(t)<0. For the above inequality, we will reduce the order of ∆^j_τu(t1+ sτ) by rewriting it as ∆^j−1_τ u(t+ (s+ 1)τ)−∆^j−1_τ u(t+sτ) forj = 1,2, . . . , n−1.

Taking into account the fact that all even terms are positive and all odd terms are negative, we will write off all the negative terms from the left hand side of this inequality. It yields

u(t₁+sτ)> 1−p^k0

p^k0(1−p)u(g(t₁+nτ) +k₀r)

n

X

i=s

¯

q(t₁+iτ).

Sinceg(t1+nτ) +k0r≤t1+m1τ and uis decreasing, by takings=m1, we obtain u(t1+m1τ)> u(t1+m1τ) 1−p^k0

p^k0(1−p)

n

X

i=m1

¯

q(t1+iτ), i.e.,

n

X

i=m1

¯

q(t1+iτ)< p^k0(1−p) 1−p^k0 .

This inequality contradicts (2.3). If x(t) is an eventually negative solution such that y(t) is not oscillatory, then y(t) >0 holds eventually. The above reasoning with an obvious minor modification also leads to a contradiction. Therefore, for every solutionx(t), eitherx(t) ory(t) is oscillatory.

Proof of Corollary 2.3. Without loss of generality, we suppose (1.1) has an eventu- ally positive solutionx(t) such that y(t) =x(t)−px(t−r) is not oscillatory. The proof is the same as that of Theorem 2.2 up to (5.4). By the same technique we reduce the order of the difference on the left hand side of this inequality down to the second order and it yields

∆²_τu(t1+sτ)> 1−p^k0

p^k0(1−p)u(g(t1+nτ) +k0r)

n

X

i=s

¯

q(t1+iτ).

Summing the above inequality forsfromm1 ton, we have

∆τu(t1+(n+1)τ)−∆τu(t1+m1τ)> 1−p^k0

p^k0(1−p)u(g(t1+nτ)+k0r)

n

X

s=m1

n

X

i=s

¯

q(t1+iτ).

Due to ∆_τu(t)<0, it follows from the above inequality that

−∆_τu(t₁+m₁τ)> 1−p^k0

p^k0(1−p)u(g(t₁+nτ) +k₀r)

n

X

s=m1

n

X

i=s

¯

q(t₁+iτ),

so

u(t₁+m₁τ)> 1−p^k0

p^k0(1−p)u(g(t₁+nτ) +k₀r)

n

X

s=m1

n

X

i=s

¯

q(t₁+iτ).

According tog(t1+nτ) +k0r≤t1+m1τ anduis decreasing, it follows that u(t₁+m₁τ)> 1−p^k0

p^k0(1−p)u(t₁+m₁τ)

n

X

i=m₁

(i−m₁+ 1)¯q(t₁+iτ), i.e.,

n

X

i=m₁

(i−m1+ 1)¯q(t1+iτ)< p^k0(1−p) 1−p^k0 .

This inequality contradicts (2.4). Thus conclusion holds.

Proof of Corollary 2.5. The proof is the same as that of Theorem 2.2 up to (5.4).

We reduce the order of the difference at the left hand of this inequality down to thelth order as we did in the proof of Theorem 2.2. Since 1≤l < m,l is odd, we obtain

−∆^l_τu(t₁+sτ)> 1−p^k0

p^k0(1−p)u(g(t₁+nτ) +k₀r)

n

X

i=s

¯

q(t₁+iτ), and iflis even,

∆^l_τu(t1+sτ)> 1−p^k0

p^k0(1−p)u(g(t1+nτ) +k0r)

n

X

i=s

¯

q(t1+iτ).

We can reach the same conclusion for the above two cases. Thus, we only give the details of the proof whenl is odd. Summing up the above inequality forsfromm_l ton, we have

−∆^l−1_τ u(t1+ (n+ 1)τ) + ∆^l−1_τ u(t1+mlτ)

> 1−p^k0

p^k0(1−p)u(g(t1+nτ) +k0r)

n

X

s=m_l n

X

i=s

¯

q(t1+iτ).

Since ∆^l−1_τ u(t)>0, the above inequality implies

∆^l−1_τ u(t₁+m_lτ)> 1−p^k0

p^k0(1−p)u(g(t₁+nτ) +k₀r)

n

X

s=ml

n

X

i=s

¯

q(t₁+iτ).

By repeating the above procedure, we obtain u(t1+m1τ)> 1−p^k0

p^k0(1−p)u(g(t1+nτ) +k0r)

n

X

m₂=m₁ n

X

m₃=m₂

· · ·

n

X

s=m_l n

X

i=s

¯

q(t1+iτ).

Becauseg(t1+nτ) +k0r≤t1+m1τ anduis decreasing, we have 1> 1−p^k0

p^k0(1−p)

n

X

m2=m1

n

X

m3=m2

· · ·

n

X

s=ml

n

X

i=s

¯

q(t₁+iτ)

= 1−p^k0 p^k0(1−p)

Xⁿ

m2=m1

n

X

m3=m2

. . .

×

n

X

i=ml−1

1

2!(i−m_l−1+ 1)(i−m_l−1+ 2)¯q(t₁+iτ) . . .

= 1−p^k0 p^k0(1−p)

Xⁿ

i=m1

¯

q(t1+iτ)

i

X

m₂=m₁

1

(l−1)!(i−m2+ 1)(i−m2+ 2)

×. . .(i−m2+ (l−1))

= 1−p^k0 p^k0(1−p)

Xⁿ

i=m1

1

l!(i−m1+ 1)(i−m1+ 2)×. . .(i−m1+l)¯q(t1+iτ) , i.e.,

1 l!

n

X

i=m₁

(i−m1+ 1). . .(i−m1+l)¯q(t1+iτ)<p^k0(1−p) 1−p^k0 .

This inequality contradicts (2.5). Thus the conclusion holds.

Proof of Theorem 2.7. The proof is similar to that of Theorem 2.2. However, the proof of the feature ofz(t) is different from that of Theorem 2.2 due top= 1. We, hence, just give the proof about the feature of z(t). For z(t), by Lemma 4.2, we noticehcould be even with 2≤h≤m−2. So ∆ⁱ_τz(t)>0 for all even number i with 2≤i≤m−2, and|∆^j_τz(t)|>0 for all odd numberj with 1≤j≤m−1.

If ∆_τz(t)>0, from the proof of Theorem 2.2 we havez(t₁+dτ)→+∞asd→ ∞ for somet₁≥t₀. Sincep= 1, from 0< x(t)< x(t−r), we know thatx(t) is bounded on [t₀,∞). Thus, z(t) is bounded on [t₀,∞). This contradictsz(t₁+dτ)→+∞

as d→ ∞. Thus, ∆τz(t)<0. So, according to Lemma 4.2 again, h= 0. Thus,

∆ⁱ_τz(t)>0 for all even number iwith 2≤i≤m−2 and ∆^j_τz(t)<0 for all odd numberj with 1≤j≤m−1.

The rest of the proof is as in Theorem 2.2, replacing p^k0(1−p)/(1−p^k0) by

1/k0.

The proofs of the following corollaries are very similar to those of Corollaries 2.3-2.5 except minor changes. Thus, we omit them.

Proof of Theorem 2.10. Suppose thatx(t) is a bounded eventually positive solution of (2.2). The proof of Theorem 2.2 is then still valid for Theorem 2.10 subject to a few obvious minor changes. Therefore, we omit the proof of the results following

equation (1.1) withp >1.

Acknowledgements. S. Wu acknowledges support from the NNSF of China via Grant 11171306 and the interdisciplinary research funding from Zhejiang University of Science and Technology via Grant 2012JC09Y.

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Shuhui Wu (corresponding author)

School of Science, Zhejiang University of Science and Technology, 318 Liuhe Road, Hangzhou 310023, China

E-mail address:[email protected], [email protected]

Pargat Singh Calay

Faculty of Computing, London Metropolitan University, 166–220 Holloway Road, Lon- don N7 8DB, UK

E-mail address:[email protected]

Zhanyuan Hou

Faculty of Computing, London Metropolitan University, 166–220 Holloway Road, Lon- don N7 8DB, UK

E-mail address:[email protected]